Write epsilon-delta proof to prove that lim(x->2) 4/x^2 = 1?

Scratchwork:

Let e > 0, and assume |4/x^2 -1| < e

then | (4-x^2)/x^2| < e

or | (x-2)(x+2)/x^2 | < e

Suppose now that |x-2| < 1 so that -1 < x-2 < 1 or 1 < x < 3

An upper bound of (x+2)/x^2 = 1/x + 2/x^2 over (1,3) is when x=1

and this is 1/1+2/1 = 3

This tells us that if | (x-2)(x+2)/x^2 | < 3| (x-2)|

and we make 3|x-2| < e then |x-2| < e/3

So we set d = min(1, e/3)

Actual Proof:

Given e > 0, set d = min(1,e/3)

Let x be such that |x-2| < d

Then |4/x^2 - 1| = |(4-x^2)/x^2| = |(x-2)(x+2)/x^2| < |x-2| * 3

( where i have used when |x-2| < 1 that (x+2)/x^2| < 3 as shown above)

and since |x-2| < e/3 then it follows that

|4/x^2 - 1| < |x-2| * 3 < e/3*3 = e

We have proved that if x such that |x-2| < d then |4/x^2 - 1| < e

where d = min(1,e/3)

or we have proved lim(x->2) 4/x^2 = 1

Given ε > 0, we need to find δ > 0 such that

0 < |x - 2| < δ ==> |4/x² - 1| < ε.

To this end, note that

|4/x² - 1|

= |(4 - x²)/x²|

= |x² - 4|/|x²|

= |x - 2| |x + 2|/x²

< |x - 2| |x + 2|/1², assuming that |x - 2| < 1 (so that 1 < x < 3 and x is away from 0)

= |x - 2| |(x - 2) + 1|

≤ |x - 2| [|x - 2| + |1|], by triangle inequality

= |x - 2|² + |x - 2|

< |x - 2| + |x - 2|, assuming that |x - 2| < 1 (as above)

= 2|x - 2|.

So given ε > 0, let δ = min{1, ε/2}. Then, 0 < |x - 2| < δ implies that

|4/x² - 1| < 2|x - 2| < 2(ε/2) = ε, as required.

I hope this helps!